<p>If \(a, b, c \in R^+\), then the minimum value of \(a(b^2 + c^2) + b(c^2 + a^2) + c(a^2 + b^2)\) is equal to</p>
Step-by-Step Solution
Key Concept: Expand the expression symmetrically and apply AM-GM inequality to recognize that the minimum occurs when variables are equal (a = b = c), which yields 6a³ when evaluated at the critical point.
<p><strong>Step 1:</strong> Expand the expression:</p><p>f(a,b,c) = ab² + ac² + bc² + ba² + ca² + cb²</p><p>= ab² + ba² + bc² + cb² + ca² + ac²</p><p>= ab(a+b) + bc(b+c) + ca(c+a)</p><p><strong>Step 2:</strong> Recognize the symmetric nature. Rearrange as:</p><p>f(a,b,c) = a²(b+c) + b²(a+c) + c²(a+b)</p><p><strong>Step 3:</strong> Apply calculus or AM-GM principle. By symmetry and convexity, the extremum occurs at a = b = c.</p><p><strong>Step 4:</strong> Substitute a = b = c:</p><p>f(a,a,a) = a(a² + a²) + a(a² + a²) + a(a² + a²)</p><p>= 3a · 2a² = 6a³</p><p><strong>Step 5:</strong> Since a, b, c ∈ ℝ⁺ and the function grows without bound, we examine the normalized case. When a = b = c = 1:</p><p>f(1,1,1) = 1(1+1) + 1(1+1) + 1(1+1) = 2 + 2 + 2 = 6</p><p>∴ <strong>Answer: D (The minimum value is 6abc or 6 when normalized)</strong></p>
Correct Answer: D